Problem: Let $R$ be the region in the first and second quadrants enclosed by the polar curve $r(\theta)=\theta+\sin(2\theta)$ and the $x$ -axis, as shown in the graph. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{0}^{\scriptsize\dfrac{\pi}{2}}\left(\dfrac{1}{2}\theta^2+\theta\sin(2\theta)+\dfrac{1}{2}\sin^2(2\theta)\right)d\theta$ (Choice B) B $ \int_{0}^{\pi}\left(\dfrac{1}{2}\theta^2+\theta\sin(2\theta)+\dfrac{1}{2}\sin^2(2\theta)\right)d\theta$ (Choice C) C $ \int_{0}^{\scriptsize\dfrac{\pi}{2}}\left(\theta^2+2\theta\sin(2\theta)+\sin^2(2\theta)\right)d\theta$ (Choice D) D $ \int_{0}^{\pi}\left(\theta^2+2\theta\sin(2\theta)+\sin^2(2\theta)\right)d\theta$
Answer: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ We know $r(\theta)$ but we still need to figure out $\alpha$ and $\beta$. Since $R$ is in enclosed by the $x$ -axis on both the first and second quadrants and $r(\theta)\geq 0$ for all non-negative values of $\theta$, its boundaries are $\alpha=0$ and $\beta=\pi$. Let's plug ${r(\theta)=\theta+\sin(2\theta)}$, ${\alpha=0}$, and ${\beta=\pi}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{0}}^{{\pi}}\dfrac{1}{2}\left({\theta+\sin(2\theta)}\right)^{2}d\theta \\\\ &= \int_{0}^{\pi}\dfrac{1}{2}\left(\theta^2+2\theta\sin(2\theta)+\sin^2(2\theta)\right)d\theta \\\\ &= \int_{0}^{\pi}\left(\dfrac{1}{2}\theta^2+\theta\sin(2\theta)+\dfrac{1}{2}\sin^2(2\theta)\right)d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{0}^{\pi}\left(\dfrac{1}{2}\theta^2+\theta\sin(2\theta)+\dfrac{1}{2}\sin^2(2\theta)\right)d\theta$